Five resistor each having value of 4 `Omega` are connected with ideal battery and ammeter as shown in figure. Find reading of ammeter.

Explanation

When 4 resistances each of value `4 Omega` are connected in parallel equivalent resistance would be `(R)/(n) = (4)/(4) = 1 Omega`. Hence , above circuit can be redrawn as follows :

Currne in the ammneter in above closed loop is ,
`I = (epsilon )/(R_(eq)) = (2)/(4 + 1 ) = (2)/(5) = 0.4 ` A

Equivalent resictance of pair of resistors, ` = (4 xx 4)/(4 + 4 ) = (16)/(8) = 2 Omega`
Equivalent resistance of circuit ,
`R = (2 xx 2 )/(2 + 2 ) + 4 `
= 1 + 4
= 5 `Omega`
`therefore ` Current in ammeter I ` = (epsilon)/(R)`
`I = (2)/(3) `
I = 0.4 A